Problem: $\sum\limits_{n=1}^{\infty } \dfrac{(n+6)(x+4)^n}{n\cdot7^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-10 \le x<2$ (Choice B) B $-10<x<2$ (Choice C) C $-11<x<3$ (Choice D) D $-11\le x<3$
We use the ratio test. For $x\neq -4$, let $a_n= \dfrac{(n+6)(x+4)^n}{n\cdot7^n} $. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x+4}{7}\right| $ The series converges when $\left| \dfrac{x+4}{7}\right| <1$, which is when $-11<x<3$. Now let's check the endpoints, $x=-11$ and $x=3$. Letting $x=-11$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(n+6)(-11+4)^n}{n\cdot7^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(n+6)(-7)^n}{n\cdot7^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot(n+6)\cdot 7^n}{n\cdot7^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot(n+6)}{n} \end{aligned}$ By the $n^{\text{th}}$ -term test, we know this series diverges. Letting $x=3$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(n+6)(3+4)^n}{n\cdot7^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(n+6)7^n}{n\cdot7^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(n+6)}{n} \end{aligned}$ By the $n^{\text{th}}$ -term test, we know this series diverges. In conclusion, the interval of convergence is $-11<x<3$.